Loading data

library(ggplot2)
library(tidyverse)
library(kableExtra)

who <- read.csv("who.csv", header = T)

kable(head(who)) %>%
  kable_styling(bootstrap_options = c("striped", "hover", "condensed"), full_width = F)
Country LifeExp InfantSurvival Under5Survival TBFree PropMD PropRN PersExp GovtExp TotExp
Afghanistan 42 0.835 0.743 0.99769 0.0002288 0.0005723 20 92 112
Albania 71 0.985 0.983 0.99974 0.0011431 0.0046144 169 3128 3297
Algeria 71 0.967 0.962 0.99944 0.0010605 0.0020914 108 5184 5292
Andorra 82 0.997 0.996 0.99983 0.0032973 0.0035000 2589 169725 172314
Angola 41 0.846 0.740 0.99656 0.0000704 0.0011462 36 1620 1656
Antigua and Barbuda 73 0.990 0.989 0.99991 0.0001429 0.0027738 503 12543 13046

##Question 1

Provide a scatterplot of LifeExp ~ TotExp, and run simple linear regression. Do not transform the variables. Provide and interpret the F statistics, R2, standard error,and p-values only. Discuss whether the assumptions of simple linear regression met.

Scatterplot

qplot(who$TotExp, who$LifeExp, main="Total Country Healthcare Expenditures vs. Life Expectancy", xlab = "Total Country Expenditures on Healthcare (USD)", ylab = "Avg. Life Expectancy (Years)") + 
  scale_x_continuous(labels = function(x) format(x, scientific = FALSE))

Model

model_who <- lm(LifeExp~TotExp,who)
plot(model_who)

summary(model_who)
## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = who)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -24.764  -4.778   3.154   7.116  13.292 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 6.475e+01  7.535e-01  85.933  < 2e-16 ***
## TotExp      6.297e-05  7.795e-06   8.079 7.71e-14 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 9.371 on 188 degrees of freedom
## Multiple R-squared:  0.2577, Adjusted R-squared:  0.2537 
## F-statistic: 65.26 on 1 and 188 DF,  p-value: 7.714e-14
model_who
## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = who)
## 
## Coefficients:
## (Intercept)       TotExp  
##   6.475e+01    6.297e-05

interpretation

Question 2

Raise life expectancy to the 4.6 power (i.e., LifeExp^4.6). Raise total expenditures to the 0.06 power (nearly a log transform, TotExp^.06). Plot \(LifeExp^4.6\) as a function of \(TotExp^.06\), and r re-run the simple regression model using the transformed variables. Provide and interpret the F statistics, \(R^2\), standard error, and p-values. Which model is “better?”

life_exp <- who$LifeExp^4.6
tot_exp <- who$TotExp^0.06

Scatterplot

qplot(tot_exp, life_exp, main="Total Country Healthcare Expenditures vs. Life Expectancy (Transformed)", xlab = "Total Country Expenditures on Healthcare (USD^0.06)", ylab = "Avg. Life Expectancy (Years^4.6)") + geom_smooth(method="lm")  + 
  scale_y_continuous(labels = function(x) format(x, scientific = FALSE))

Model

model_who_2 <- lm(life_exp ~ tot_exp)
summary(model_who_2)
## 
## Call:
## lm(formula = life_exp ~ tot_exp)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -308616089  -53978977   13697187   59139231  211951764 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -736527910   46817945  -15.73   <2e-16 ***
## tot_exp      620060216   27518940   22.53   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 90490000 on 188 degrees of freedom
## Multiple R-squared:  0.7298, Adjusted R-squared:  0.7283 
## F-statistic: 507.7 on 1 and 188 DF,  p-value: < 2.2e-16
model_who_2
## 
## Call:
## lm(formula = life_exp ~ tot_exp)
## 
## Coefficients:
## (Intercept)      tot_exp  
##  -736527909    620060216

interpretation

  • R-squared: An adjusted R2 of about 0.73 means that healthcare expenditure per country explains 73% of a country’s life expectancy. This is much higher than the previous model.

  • Standard error: The model has a standard error of 46,817,945 for the intercept and 27,518,940 for the slope. These are pretty big errors relative to the coefficients, which suggests that the model is not estimating them very well.

  • P-Value: The model’s p-value of 2e−16 is tiny. This suggests that there is a high likelihood that total healthcare expenditure is relevant in the model, and the model more accurately predicts it. However, this doesn’t align with the model’s other parameters.

  • F-Statistic: The F-test reports the p-value of a model that has one fewer parameters than the current one. Since we are only considering one variable in this model, the F-statistic is not meaningful here.

The second model has a higher \(R^2\) at the same level of significance, which suggests that it is better at explaining the variance in life expectancy. However, it also has a bigger margin of error, adding more uncertainty to the predicted values.

Question 3

Using the results from 2, forecast life expectancy when \(TotExp^.06 = 1.5\). Then forecast life expectancy when \(TotExp^.06 = 2.5\).

forecasted_Life_expectancy <- function(x){
  ans <- model_who_2$coefficients[1] + model_who_2$coefficients[2] * x
  return(ans ** (1/4.6)) # Must transform the data into a sensible number
}
print(paste0("The forecasted life expectancy when TotExp^.06 = 1.5 is: ", round(forecasted_Life_expectancy(1.5),2), " years."))
## [1] "The forecasted life expectancy when TotExp^.06 = 1.5 is: 63.31 years."
print(paste0("The forecasted life expectancy when TotExp^.06 = 2.5 is : ", round(forecasted_Life_expectancy(2.5),2), " years."))
## [1] "The forecasted life expectancy when TotExp^.06 = 2.5 is : 86.51 years."

Question 4

Build the following multiple regression model and interpret the F Statistics, R2, standard error, and p-values. How good is the model?

model_who_4 <- lm(LifeExp ~ PropMD + TotExp + (PropMD * TotExp), who)

summary(model_who_4)
## 
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + (PropMD * TotExp), data = who)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -27.320  -4.132   2.098   6.540  13.074 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    6.277e+01  7.956e-01  78.899  < 2e-16 ***
## PropMD         1.497e+03  2.788e+02   5.371 2.32e-07 ***
## TotExp         7.233e-05  8.982e-06   8.053 9.39e-14 ***
## PropMD:TotExp -6.026e-03  1.472e-03  -4.093 6.35e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.765 on 186 degrees of freedom
## Multiple R-squared:  0.3574, Adjusted R-squared:  0.3471 
## F-statistic: 34.49 on 3 and 186 DF,  p-value: < 2.2e-16

interpretation

This model has an F-statistic of 34.49 and a statistically significant p-value < 2.2e-16, a residual standard error of 8.765, and adjusted R-squared 0.3471. This model is better than the first model(model_who) but worse than the second model(model_who_2). The only thing improved here compared to the first model is that the precision is slightly better with the standard error being at 8.765. The adjusted R-squared suggests that 34.71% of the variation in this data could be interpreted from this model.

Therefore this model is not as good as the second model.

Question 5

Forecast LifeExp when PropMD = 0.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?

forecast_life_expectancy <- function(PropMD, TotExp){
  d <- data.frame(PropMD = c(PropMD), TotExp = c(TotExp))
  prediction <- predict(model_who_4, d)
  cat(paste("Life Expectancy:", round(unname(prediction),1), "years"))
}
forecast_life_expectancy(0.03, 14)
## Life Expectancy: 107.7 years

This model assumes that 3% of the population of a country are doctors (PropMD) and the country’s total expenditure on healthcare is 14 USD (TotExp). The resulting predicted life expectancy is 107 years.

This doesn’t make sense for a lot of reasons – the total expenditure on healthcare is absurdly low at 14 USD, and the average life expectancy is absurdly high as above 100 years old.